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\newtheorem{theorem}{Theorem}[section]%定理
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\title{\heiti\zihao{2} 习题17.1}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{计算 $I=\int\limits_{L} y \mathrm{~d} s$, 其中 $L$ 为摆线 $x=a(t-\sin t), y=a(1-\cos t)(0 \leqslant t \leqslant 2 \pi)$ 的一拱.}
\textbf{解}\quad
$$
\begin{aligned}
    \dfrac{\mathrm{d}x}{\mathrm{d}t} &= a(1-\cos t)\\
    \dfrac{\mathrm{d}y}{\mathrm{d}t} &= a\sin t\\
    I=\int\limits_Ly\mathrm{d}s&=\int_0^{2\pi}a(1-\cos t)\cdot a\sqrt{(1-\cos t)^2+\sin^2t}\mathrm{d}t\\
    &=\sqrt{2}a^2\int_0^{2\pi}(1-\cos t)\sqrt{1-\cos t}\mathrm{d}t\\
    &=8a^2\int_0^{2\pi}\sin^3\dfrac{t}{2}\mathrm{d}\dfrac{t}{2}\\
    &=16a^2\int_0^{\pi/2}\sin^3 u \mathrm{d}u\\
    &=16a^2\cdot\dfrac{2!!}{3!!}=\dfrac{32a^2}{3}
\end{aligned}
$$

\section{计算 $I=\int\limits_{L}(x+y) \mathrm{d} s$, 其中 $L$ 是以点 $O(0,0), A(1,0), B(1,1)$ 为顶点的三角形的边界.}
\textbf{解}\quad
由第一型曲线积分的性质,可将积分曲线拆为三段:$L_1:(0,0)--(1,1)$, $L_2:(0,0)--(1,0)$, $L_3:(1,0)--(1,1)$.
在这三条曲线上,分别有
$$
\begin{array}{ccc}
    \left\{\begin{array}{l}
        x(t) = t\\
        y(t) = t
    \end{array}\right. & \left\{\begin{array}{l}
        x(t) = t\\
        y(t) = 0
    \end{array}\right. & \left\{\begin{array}{l}
        x(t) = 1\\
        y(t) = t
    \end{array}\right.
\end{array}
$$
所以有
$$
\begin{aligned}
    \int\limits_L(x+y)\mathrm{d}s &=\int\limits_{L1}(x+y)\mathrm{d}s+\int\limits_{L2}(x+y)\mathrm{d}s+\int\limits_{L3}(x+y)\mathrm{d}s\\
    &=\int\limits_0^12t\sqrt{2}\mathrm{d}t+\int\limits_0^1t\mathrm{d}t+\int\limits_0^11+t\mathrm{d}t\\
    &=\sqrt{2}+\dfrac{1}{2}+1+\dfrac{1}{2}\\
    &=2+\sqrt{2}
\end{aligned}
$$

\section{计算 $I=\int\limits_{L} x^{2} \mathrm{~d} s$, 其中 $L$ 为圆周 $\left\{\begin{array}{l}x^{2}+y^{2}+z^{2}=a^{2}\\ x+y+z=0 \end{array}\right.$}
\textbf{解}\quad
由于$L$关于$x,y,z$轮换对称,所以$\int\limits_{L} x^{2} \mathrm{~d} s = \int\limits_{L} \dfrac{x^2+y^2+z^2}{3} \mathrm{~d} s= \int\limits_{L} \dfrac{a^2}{3} \mathrm{~d} s$.所以
$$
\begin{aligned}
    I&=2\pi a\cdot \dfrac{a^2}{3}=\dfrac{2\pi a^3}{3}
\end{aligned}
$$


\section{计算 $I=\int\limits_{L} 2 x \mathrm{~d} s$, 其中 $L$ 由 $L_{1}$ 和 $L_{2}$ 连结而成, $L_{1}$ 是从 $(0,0)$ 到 $(1,1)$ 的抛物线 $y=x^{2}$,$L_{2}$ 是从 $(1,1)$ 到$(1,2)$ 的竖直线段.}
\textbf{解}\quad
$$
\begin{aligned}
    I&=\int\limits_{L1}2x\mathrm{d}s+\int\limits_{L2}2x\mathrm{d}s\\
    &=\int_0^12x\sqrt{1+(2x)^2}\mathrm{d}x+\int_1^22\mathrm{d}y\\
    &=\dfrac{5\sqrt{5}-1}{6}+2\\
    &=\dfrac{5\sqrt{5}+11}{6}
\end{aligned}
$$


\section{求曲线形物体 $x^{2}+y^{2}=2 y(y \leqslant 1)$ 对 $x$ 轴、 $y$ 轴的转动惯量,假设它的线密度为 $\rho(x, y)=1$.}
\textbf{解}\quad
$x^2+(y-1)^2 = 1$.

对$x$轴:
$$
\begin{aligned}
    \int\limits_{L}y^2\cdot 1\mathrm{d}s&=\int_\pi^{2\pi}(1-\sin\theta)^2\mathrm{d}\theta\\
    &=\dfrac{3\pi}{2}+4
\end{aligned}
$$
对$y$轴:
$$
\begin{aligned}
    \int\limits_{L}x^2\cdot 1\mathrm{d}s&=\int_pi^{2\pi}\cos^2\theta\mathrm{d}\theta\\
    &=\dfrac{\pi}{2}
\end{aligned}
$$


\section{求摆线 $\left\{\begin{array}{l}x=a(t-\sin t), \\ y=a(1-\cos t)\end{array}(0 \leqslant t \leqslant \pi)\right.$ 的质心,设其质量分布是均匀的.}
\textbf{解}\quad
总质量:
$$
\begin{aligned}
    \int\limits_{L}\sqrt{x_t^2+y_t^2}\cdot\lambda\mathrm{d}s&=\int_0^\pi a\sqrt{(1-\cos t)^2+\sin^2t}\mathrm{d}t\\
    &=4\lambda a\int_0^\pi\sin\dfrac{t}{2}\mathrm{d}\dfrac{t}{2}\\
    &=4\lambda a
\end{aligned}
$$
$x$分量:
$$
\begin{aligned}
    \int\limits_Lx\sqrt{x_t^2+y_t^2}\lambda\mathrm{d}s&=4\lambda a^2\int_0^\pi\sin\dfrac{t}{2}(t-\sin t)\mathrm{d}\dfrac{t}{2}\\
    &=4\lambda a^2\int_0^{\pi/2}\sin u(2u-\sin 2u)\mathrm{d}u\\
    &=\dfrac{16\lambda a^2}{3}
\end{aligned}
$$
$y$分量:
$$
\begin{aligned}
    \int\limits_Ly\sqrt{x_t^2+y_t^2}\lambda\mathrm{d}s&=4\lambda a^2\int_0^\pi(1-\cos t)\sin\dfrac{t}{2}\mathrm{d}t\\
    &=\dfrac{16\lambda a^2}{3}
\end{aligned}
$$
故$(\bar{x},\bar{y})=\left(\dfrac{\dfrac{16\lambda a^2}{3}}{4\lambda a},\dfrac{\dfrac{16\lambda a^2}{3}}{4\lambda a}\right)=\left(\dfrac{4a}{3},\dfrac{4a}{3}\right)$

\section{设曲线形物体 $L:\left\{\begin{array}{l}x^{2}+y^{2}=a^{2}, \\ z=0\end{array}(x \geqslant 0, y \geqslant 0)\right.$, 其线密度为 \\$\rho(x, y, z)=x$, 求它对点$M(0,0, a)$ 处的单位质点的引力 $F(a>0) .$}
\textbf{解}\quad
由对称性可知
$$
\begin{aligned}
    \int\limits_L\vec{F}\mathrm{d}l&=\int_0^{\pi/2}\dfrac{Ga^2\cos\theta\cos\theta\mathrm{d}\theta}{2\sqrt{2}a^2}\vec{i} + \int_0^{\pi/2}\dfrac{Ga^2\cos\theta\sin\theta\mathrm{d}\theta}{2\sqrt{2}a^2}\vec{j} - \int_0^{\pi/2}\dfrac{Ga^2\cos\theta\mathrm{d}\theta}{2\sqrt{2}a^2}\vec{k}\\
    &=\dfrac{G}{2\sqrt{2}}\int_0^{\pi/2}(1-\sin^2\theta)\mathrm{d}\theta\vec{i} + \dfrac{G}{2\sqrt{2}}\int_0^{\pi/2}\cos\theta\sin\theta\mathrm{d}\theta\vec{j}-\dfrac{G}{2\sqrt{2}}\int_0^{\pi/2}\cos\theta\mathrm{d}\theta\vec{k}\\
    &=\dfrac{G\pi}{8\sqrt{2}}\vec{i}+\dfrac{G}{4\sqrt{2}}\vec{j}-\dfrac{G}{2\sqrt{2}}\vec{k}
\end{aligned}
$$
其中$\vec{i},\vec{j}$为$x,y$方向的单位向量.

\section{证明:若函数 $f(x, y)$ 在光滑曲线 $L: x=x(t), y=y(t), t \in[\alpha, \beta]$ 上连续,则存在点 $\left(x_{0}, y_{0}\right) \in L$, 使得 $\int\limits_{l} f(x, y) \mathrm{d} s=f\left(x_{0}, y_{0}\right) l$, 其中 $l$ 为 $L$ 的弧长.}
\begin{proof}
    由积分第一中值定理:
    $$
        \begin{aligned}
            \int\limits_lf(x,y)\mathrm{d}s&=\int_\alpha^\beta f(x(t),y(t))\sqrt{x^2_t+y^2_t}\mathrm{d}t\\
            &=\eta\int_\alpha^\beta\sqrt{x^2_t+y^2_t}\mathrm{d}t\\
            &=f(x_0,y_0)l
        \end{aligned}
    $$
\end{proof}

\section{求曲面 $x^{2}+y^{2}=1$ 与 $x^{2}+z^{2}=1$ 围成立体的表面积.}
\textbf{解}\quad
显然表面为上下两个牟合方盖,考虑其中一部分:$x^2+y^2 = 1, y = z$,其与$xoy$平面所夹得面积的$16$倍即为所求.记$z=f(x,y)$.
\begin{figure}[h]
    \centering
    \includegraphics[scale = 0.3]{include_picture/17.1.1.jpg}
    \caption{}
\end{figure}
$$
\begin{aligned}
    S&=16\int\limits_L z\mathrm{d}s\\
    &=16\int_0^{\pi/2}\sin t \mathrm{d}t\\
    &=16
\end{aligned}
$$
故面积为$16$.





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